What is the integral of (sqrt(x^2-9))/(x^2) ?

The integral of (sqrt(x^2-9))/(x^2) is ln(1/3 |sqrt(x^2-9)+x|)-(sqrt(x^2-9))/x+C

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integral of (sqrt(x^2-9))/(x^2)

\int \frac{x ^{2} - 9}{x ^{2}} d x

ln (\frac{1}{3} x^{2} - 9 + x) - \frac{x ^{2} - 9}{x} + C

Solution steps

\int \frac{x ^{2} - 9}{x ^{2}} d x

Apply Trigonometric Substitution

: \int tan^{2} (u) cos (u) d u

= \int tan^{2} (u) cos (u) d u

Simplify

tan^{2} (u) cos (u) : sin (u) tan (u)

= \int sin (u) tan (u) d u

Rewrite using trig identities

= \int \frac{1 - cos ^{2} ( u )}{cos ( u )} d u

Expand

\frac{1 - cos ^{2} ( u )}{cos ( u )} : \frac{1}{cos ( u )} - cos (u)

= \int \frac{1}{cos ( u )} - cos (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{cos ( u )} d u - \int cos (u) d u

\int \frac{1}{cos ( u )} d u = ln ∣ tan (u) + sec (u) ∣

\int cos (u) d u = sin (u)

= ln ∣ tan (u) + sec (u) ∣ - sin (u)

Substitute back

u = arcsec (\frac{1}{3} x)

= ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)) - sin (arcsec (\frac{1}{3} x))

Simplify

ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)) - sin (arcsec (\frac{1}{3} x)) : ln (\frac{1}{3} x^{2} - 9 + x) - \frac{x ^{2} - 9}{x}

= ln (\frac{1}{3} x^{2} - 9 + x) - \frac{x ^{2} - 9}{x}

Add a constant to the solution

= ln (\frac{1}{3} x^{2} - 9 + x) - \frac{x ^{2} - 9}{x} + C

y^{^{''}} - (\frac{6}{x}) y^{^{'}} - (\frac{4}{x ^{2}}) y = 0

\frac{d}{d a} (\frac{π ( 9 a ^{2} + 6 a + 1 )}{36 a})

n = 0 \sum \infty 1.0 5^{n}

\frac{d}{d x} (π sin (π x))

\frac{0.26 s}{( s + 1 )}

What is the integral of (sqrt(x^2-9))/(x^2) ?
The integral of (sqrt(x^2-9))/(x^2) is ln(1/3 |sqrt(x^2-9)+x|)-(sqrt(x^2-9))/x+C